Polynomials

Yes, zero is a rational number. It can be written as:
0 = 0/1 = 0/2 = 0/3 = 0/(-5) etc.
In all these cases, p = 0 (integer) and q is a non-zero integer. Therefore zero is a rational number.

Multiply 3 and 4 by (6+1) = 7: 3 = 21/7 and 4 = 28/7.
Six rational numbers between 21/7 and 28/7:
22/7, 23/7, 24/7, 25/7, 26/7, 27/7.

Multiply numerator and denominator by 6: 3/5 = 18/30 and 4/5 = 24/30.
Five rational numbers between them:
19/30, 20/30, 21/30, 22/30, 23/30.
Simplified: 19/30, 2/3, 7/10, 11/15, 23/30.

(i) True. The set of natural numbers N = {1, 2, 3, ...} is a subset of whole numbers W = {0, 1, 2, 3, ...}. Every natural number is included in the whole numbers.

(ii) False. Negative integers (like -1, -2, -3, ...) are integers but not whole numbers. Whole numbers do not include negatives.

(iii) False. Rational numbers like 1/2, 3/4, -2/5 are rational but not whole numbers. Whole numbers are specific non-negative integers only.

(i) True. Real numbers include both rational and irrational numbers. So every irrational number is certainly a real number.

(ii) False. Points on the number line include negative numbers and zero. Negative numbers cannot be expressed as √m where m is a natural number (as √m is always non-negative).

(iii) False. Real numbers include rational numbers too (like 2, 3/4, -5). Not every real number is irrational.

No, not all square roots of positive integers are irrational. For example:
√4 = 2 (rational), √9 = 3 (rational), √16 = 4 (rational).
Square roots of perfect squares are rational numbers.

Steps to represent √5 on the number line:
1. On the number line, mark point O at origin (0) and point A at 2 (so OA = 2 units).
2. At A, draw AB perpendicular to OA, with AB = 1 unit.
3. Join OB. By Pythagoras: OB = √(OA² + AB²) = √(4 + 1) = √5.
4. With O as center and OB as radius, draw an arc intersecting the number line at C.
5. OC = OB = √5. The point C represents √5 on the number line.

(i) 36/100 = 0.36 — Terminating decimal.

(ii) 1/11 = 0.090909... = 0.09̅ — Non-terminating repeating.

(iii) 4(1/8) = 33/8 = 4.125 — Terminating decimal.

(iv) 3/13 = 0.230769230769... = 0.230769̅ — Non-terminating repeating.

(v) 2/11 = 0.1818... = 0.18̅ — Non-terminating repeating.

(vi) 329/400 = 0.8225 — Terminating decimal.

Since 1/7 = 0.142857̅, multiply by 2, 3, 4, 5, 6 respectively (using cyclic pattern of 142857):
2/7 = 0.285714̅
3/7 = 0.428571̅
4/7 = 0.571428̅
5/7 = 0.714285̅
6/7 = 0.857142̅

(i) Let x = 0.6̅ = 0.6666...
10x = 6.6666...
10x - x = 6 ⇒ 9x = 6 ⇒ x = 2/3.

(ii) Let x = 0.47̅ = 0.4777...
10x = 4.777...
100x = 47.777...
100x - 10x = 43 ⇒ 90x = 43 ⇒ x = 43/90.

(iii) Let x = 0.001̅ = 0.001001...
1000x = 1.001001...
1000x - x = 1 ⇒ 999x = 1 ⇒ x = 1/999.

Let x = 0.9999...
10x = 9.9999...
10x - x = 9 ⇒ 9x = 9 ⇒ x = 1.
So 0.999... = 1. This can be surprising! It confirms that 0.9̅ is just another way to write the number 1. There is no gap between 0.999... and 1 — they are equal.

When dividing by 17, the remainder at each step can be 0, 1, 2, ..., 16 (17 possible values).
The maximum number of digits in the repeating block of 1/17 is therefore 16.
(Actual value: 1/17 = 0.0588235294117647̅ — 16 repeating digits).

A rational number p/q (in lowest terms) has a terminating decimal if and only if the prime factorisation of q has no prime factors other than 2 and 5 (i.e., q = 2^m × 5ⁿ).
Otherwise, it is a non-terminating repeating decimal.
Examples:
3/8 = 3/2³ ⇒ Terminating (0.375).
7/6 = 7/(2×3) ⇒ Non-terminating repeating (1.1666...).

Non-terminating and non-repeating decimals are irrational numbers. Three examples:
1. √2 = 1.41421356...
2. √3 = 1.73205080...
3. π = 3.14159265...

5/7 = 0.714285... and 9/11 = 0.818181...
Three irrational numbers between 0.714... and 0.818...:
1. 0.7201002000300004... (non-terminating, non-repeating)
2. 0.750750075000750...
3. 0.808008000800008...

(i) √23 ⇒ 23 is not a perfect square ⇒ Irrational.

(ii) √225 = 15 ⇒ Perfect square ⇒ Rational.

(iii) 0.3796 ⇒ Terminating decimal ⇒ Rational.

(iv) 7.478478... ⇒ Non-terminating but repeating (7.478̅) ⇒ Rational.

(v) 1.101001000100001... ⇒ Non-terminating and non-repeating ⇒ Irrational.

Step 1: Locate 3.765 between 3 and 4 on the number line.
Step 2: Divide the segment [3, 4] into 10 equal parts. 3.765 is between 3.7 and 3.8.
Step 3: Divide [3.7, 3.8] into 10 equal parts. 3.765 lies between 3.76 and 3.77.
Step 4: Divide [3.76, 3.77] into 10 equal parts. 3.765 is the 5th mark from 3.76.
This process of zooming in is called successive magnification.

4.26̅ = 4.2666...
Step 1: 4.2666 lies between 4 and 5. More precisely between 4.2 and 4.3.
Step 2: Divide [4.2, 4.3] into 10 parts. 4.2666 lies between 4.26 and 4.27.
Step 3: Divide [4.26, 4.27] into 10 parts. 4.2666 lies between 4.266 and 4.267.
Step 4: Divide [4.266, 4.267] into 10 parts. 4.2666 is between 4.2666 and 4.2667 at the 6th mark.
Mark this with a dot to represent 4.26̅.

(i) 2 - √5: √5 is irrational, 2 is rational. Rational - Irrational = Irrational.

(ii) (3 + √23) - √23 = 3: This simplifies to 3 ⇒ Rational.

(iii) 2√7 / 7√7 = 2/7: This simplifies to 2/7 ⇒ Rational.

(iv) 1/√2: √2 is irrational, so 1/√2 is also Irrational.

(v) 2π: π is irrational, so 2π is also Irrational.

(i) (3 + √3)(2 + √2) = 6 + 3√2 + 2√3 + √6 = 6 + 3√2 + 2√3 + √6.

(ii) (3 + √3)(3 - √3) = 3² - (√3)² = 9 - 3 = 6.

(iii) (√5 + √2)² = (√5)² + 2√5√2 + (√2)² = 5 + 2√10 + 2 = 7 + 2√10.

(iv) (√5 - √2)(√5 + √2) = (√5)² - (√2)² = 5 - 2 = 3.

When mathematicians talk about π, the value is precisely the ratio of the circumference to the diameter. However, this ratio is not always rational — it depends on whether both c and d are rational at the same time.
It has been proven that π is irrational. Its decimal value 3.14159265358979... is non-terminating and non-repeating. Therefore, π cannot be expressed as p/q.

(i) 1/√7: Multiply by √7/√7 = √7/7.

(ii) 1/(√7 - √6): Multiply by (√7 + √6)/(√7 + √6)
= (√7 + √6)/(7 - 6) = √7 + √6.

(iii) 1/(√5 + √2): Multiply by (√5 - √2)/(√5 - √2)
= (√5 - √2)/(5 - 2) = (√5 - √2)/3.

(iv) 1/(√7 - 2): Multiply by (√7 + 2)/(√7 + 2)
= (√7 + 2)/(7 - 4) = (√7 + 2)/3.

(i) 64^1/2 = √64 = 8.

(ii) 32^1/5 = ⁵√32 = ⁵√(2⁵) = 2.

(iii) 125^1/3 = ³√125 = ³√(5³) = 5.

(i) 9^3/2 = (9^1/2)³ = 3³ = 27.

(ii) 32^2/5 = (32^1/5)² = 2² = 4.

(iii) 16^3/4 = (16^1/4)³ = 2³ = 8.

(iv) 125^-1/3 = 1/125^1/3 = 1/5 = 1/5.

(i) 2^2/3 × 2^1/5 = 2^{(2/3 + 1/5)} = 2^{(10/15 + 3/15)} = 2^13/15.

(ii) (1/3³)⁷ = 1/3²¹ = 3^-21.

(iii) 11^1/2 / 11^1/4 = 11^{(1/2 - 1/4)} = 11^1/4 = ⁴√11.

(iv) 7^1/2 × 8^1/2 = (7 × 8)^1/2 = 56^1/2 = √56 = 2√14.

Both fractions have denominator 13. Between -3/13 and 9/13, the integers in the numerator range from -2 to 8.
So 10 rational numbers: -2/13, -1/13, 0/13, 1/13, 2/13, 3/13, 4/13, 5/13, 6/13, 7/13.

Let x = 2.3181818... (the bar is on "18").
1000x = 2318.1818...
10x = 23.1818...
1000x - 10x = 2295 ⇒ 990x = 2295 ⇒ x = 2295/990 = 51/22.

Using (a + b)² = a² + 2ab + b²:
= (√3)² + 2(√3)(√5) + (√5)²
= 3 + 2√15 + 5 = 8 + 2√15.

Multiply numerator and denominator by the conjugate (3 + 2√2):
= 5(3 + 2√2) / (3² - (2√2)²)
= (15 + 10√2) / (9 - 8)
= 15 + 10√2.

1/a = 1/(2 + √3) × (2 - √3)/(2 - √3) = (2 - √3)/(4 - 3) = 2 - √3.
a - 1/a = (2 + √3) - (2 - √3) = 2√3.

(81)^-1/2 = 1/√81 = 1/9.
(125)^1/3 = ³√125 = 5.
Product = (1/9) × 5 = 5/9.

Using (a + b)(a - b) = a² - b²:
= 3² - (2√2)² = 9 - 4 × 2 = 9 - 8 = 1.

Proof by contradiction:
Assume √2 is rational. Then √2 = p/q where p, q are integers with no common factor and q ≠ 0.
Squaring: 2 = p²/q² ⇒ p² = 2q².
So p² is even, meaning p itself is even. Let p = 2m.
Then (2m)² = 2q² ⇒ 4m² = 2q² ⇒ q² = 2m².
So q² is even, meaning q is even.
But now both p and q are even, contradicting that they have no common factor. Hence √2 is irrational.

= 2⁶ × 3⁴ / (3³ × 2⁴)
= 2^6-4 × 3^4-3 = 2² × 3 = 4 × 3 = 12.

= 3^3/2 × (3²)^-1/4 = 3^3/2 × 3^-1/2
= 3^{(3/2 - 1/2)} = 3¹ = 3.

Multiply by conjugate (2√3 + √5):
= (2√3 + √5) / ((2√3)² - (√5)²)
= (2√3 + √5) / (12 - 5)
= (2√3 + √5) / 7.

1/x = 1/(3 + 2√2) × (3 - 2√2)/(3 - 2√2) = (3 - 2√2)/(9 - 8) = 3 - 2√2.
x + 1/x = (3 + 2√2) + (3 - 2√2) = 6.

Compare 2^1/2 and 3^1/3. Express with common power — LCM of 2 and 3 is 6.
2^1/2 = 2^3/6 = (2³)^1/6 = 8^1/6.
3^1/3 = 3^2/6 = (3²)^1/6 = 9^1/6.
Since 9 > 8, 9^1/6 > 8^1/6.
Therefore ³√3 > √2.

To represent √9.3 on the number line using the geometric construction method:
1. Draw a line segment AB = 9.3 units.
2. Extend AB to C such that BC = 1 unit. So AC = 10.3 units.
3. Find the midpoint M of AC.
4. Draw a semicircle with M as centre and MC as radius.
5. From B, draw a perpendicular BD to AC. D meets the semicircle.
6. BD = √(AB × BC) = √(9.3 × 1) = √9.3.
7. With B as centre and BD as radius, draw an arc to meet the number line at E. BE = √9.3.

Rationalise each term:
Term 1: √6/(√2+√3) × (√3-√2)/(√3-√2) = √6(√3-√2)/(3-2) = √18 - √12 = 3√2 - 2√3.
Term 2: 3√2/(√6+√3) × (√6-√3)/(√6-√3) = 3√2(√6-√3)/(6-3) = √2(√6-√3) = √12 - √6 = 2√3 - √6.
Term 3: 4√3/(√6+√2) × (√6-√2)/(√6-√2) = 4√3(√6-√2)/(6-2) = √3(√6-√2) = √18 - √6 = 3√2 - √6.
Result = (3√2 - 2√3) + (2√3 - √6) - (3√2 - √6) = 0.